Algebraic Independence in the Grothendieck Ring of Varieties

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TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 359, Number 4, April 2007, Pages 1653–1683 S 0002-9947(06)03975-4 Article electronically published on September 19, 2006

ALGEBRAIC INDEPENDENCE IN THE GROTHENDIECK RING OF VARIETIES

N. NAUMANN

Abstract. We give suﬃcient cohomological criteria for the classes of given varieties over a ﬁeld k to be algebraically independent in the Grothendieck ring of varieties over k and construct some examples.

1. Introduction

Let k be a field. The Grothendieck ring of varieties over k, denoted K0(Vark), was defined by A. Grothendieck in [G]. The abelian group K0(Vark) is generated by the isomorphism classes [X]of separated k-schemes of finite type X/k subject to the relations [X]=[Y ]+[X − Y ] for Y ⊂ X a closed subscheme. Multiplication is given by [X1] · [X2]:=[X1 ×k X2]andmakesK0(Vark) a commutative ring with unit [Spec(k)]. By its very definition K0(Vark) is the value group of the universal Euler-Poincaré characteristic with compact support for varieties over k and is thus a fundamental invariant of the algebraic geometry over k. The most outstanding result on the structure of K0(Vark)fork of characteristic zero is a presentation of the ring K0(Vark)in terms of generators and relations anticipated by E. Looijenga and proved by his student F. Bittner [Bi]. B. Poonen showed that for k of characteristic zero a linear combination of the classes of suitable abelian varieties is a zero divisor in K0(Vark) [P], and J. Kollár has computed the subring of K0(Vark) generated by the classes of conics for suitable base fields k [Ko]. A deep problem pertaining to the structure of K0(Vark) is the rationality of motivic zeta-functions posed by M. Kapranov [K] on which there has been recent progress due to M. Larsen and V. Lunts [LL1]. This problem is intimately related to the finite-dimensionality of motives; cf. [A] for an exposition. Furthermore, the ring K0(Vark) plays a central rôle in the theory of motivic integration; cf. [Lo]. The present note is dedicated to the following problem about the structure of K0(Vark). Given varieties Xi/k, when are their classes [Xi] ∈ K0(Vark) algebraically independent (in the sense made precise at the end of this Introduction)? We give a number of sufficient cohomological conditions for this to be the case and construct some examples. Our main results are also valid in positive characteristic where there is no known Theorem about the structure of K0(Vark). We now review the content of the individual sections.

Received by the editors July 28, 2004 and, in revised form, January 20, 2005. 2000 Mathematics Subject Classiﬁcation. Primary 14A10. Key words and phrases. Grothendieck ring of varieties, motivic measure.

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In section 2 we construct various ring homomorphisms µk : K0(Vark) −→ R (so called motivic measures) all based on étale cohomology. The weight filtration is used to show that the classes of proper smooth varieties over a finitely generated field k are algebraically independent in K0(Vark) as soon as their first étale cohomology groups are algebraically independent in the ring of virtual l-adic Galois representations (Corollary 3). We then reformulate known properties of rationality and l-independence for varieties over Fq (Proposition 7). In Subsection 3.1 we reduce the problem of algebraic independence of virtual l-adic Galois representations to a problem about representations of (possibly non- connected) reductive groups, give a result illustrating the subtleties arising from this non-connectedness (Theorem 11) and construct an infinite sequence of curves over a given finite field k the classes of which are algebraically independent in K0(Vark) (Theorems 12 and 17). This shows in particular that K0(Vark) contains a polynomial ring in infinitely many variables and supports the intuition that K0(Vark) should be a large ring. We obtain a similar result for number fields using only elliptic curves (Theorem 13). In Subsection 3.2 we use a lemma of Skolem together with the aforementioned rationality properties to give another approach to the algebraic independence of virtual l-adic Galois representations and show that a generic pair of curves over a finite field has algebraically independent classes in the Grothendieck ring of varieties (Theorem 26 in Subsection 3.5). We also determine the structure of the subring

of K0(VarFq ) generated by zero-dimensional varieties showing in particular that

K0(VarFq ) contains infinitely many zero divisors (Theorem 25). The following definition will be central throughout the rest of this note. Definition 1. Let R be a commutative unitary ring and let nZ be the kernel of Z −→ R.Wecallaset{xi : i ∈ I} of elements xi ∈ R algebraically independent if the morphism (Z/n)[Ti : i ∈ I] −→ R, Ti → xi is injective. If R is a field, then this means algebraically independent over the prime field of R in the usual sense. Note that 1/2 ∈ Q is algebraically dependent though not integral over Z; cf. Theorem 24. All the rings appearing below will have n = 0, i.e. Z ⊂ R. In this case, elements xi ∈ R are algebraically independent if and only if the subring of R they generate is a polynomial ring over Z in the variables xi.

2. A motivic measure 2.1. General construction. Let k be a field. We fix a separable closure k of k and a rational prime l different from the characteristic of k.ForX/k separated and of finite type we denote by i i × Q ≥ Hc(X):=Hc(X k k, l)fori 0

the étale cohomology with compact support and constant coefficients Ql of the base × i change X k k.ThenHc(X) is naturally a Gk := Aut (k/k)-module, and denoting Q by Rep Gk l the category of finite-dimensional continuous representations of Gk over Q (i.e. l-adic Galois representations) we have a motivic measure l −→ Q → − i i (1) µk : K0(Vark) K0(Rep Gk l) , [X] ( 1) [Hc(X)]. i

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In fact, (1) is (well defined and) a homomorphism of abelian groups by excision, multiplicative by the Künneth formula and clearly preserves the unit. For a finite extension k ⊂ L ⊂ k we have a ring homomorphism (base change)

−×k L : K0(Vark) −→ K0(VarL) , [X] → [X ×k L] and a homomorphism of abelian groups (restriction of scalars)

/k : K0(VarL) −→ K0(Vark) , [X −→ Spec(L)] → [X −→ Spec(L) −→ Spec(k)].

From the inclusion GL ⊂ Gk we have an exact tensor functor (restriction) (2) ResGk :Rep Q −→ Rep Q GL Gk l GL l and an exact functor (induction) G (3) Ind k :Rep Q −→ Rep Q ,V→ Q [G ] ⊗Q V GL GL l Gk l l k l[GL] inducing a ring homomorphism and a homomorphism of abelian groups on the level of K0 to be denoted by the same symbols. The following diagram is commutative:

µk / K (Var ) K0(Rep Ql) 0 Ok GOk

Gk Gk −×kL /k Res Ind GL GL µL / Q K0(VarL) K0(RepGL l)

2.2. Weight filtration. Notation being as in Subsection 2.1 we now assume in addition that k is finitely generated. In this subsection we will incorporate the ∗ weight filtration on Hc (X) into the motivic measure (1). We need to assume that k is finitely generated in order to have a theory of weights. Vaguely speaking, this is a substitute (valid in any characteristic) of Hodge theory and will allow us to isolate H1 from the Euler-Poincaré characteristic in Theorem 2 below.

We refer the reader to [J], 6, for the definition of the full subcategory WRep Q ⊂ Rep Q Gk l Gk l of l-adic Galois representations having a weight filtration. From [J], Lemma 6.8.2 and (6.8.3), we know that Hi(X) ∈ WRep Q holds for all X/k separated and of c Gk l finite type and hence (1) factors as (4) µ : K (Var ) −→ K (WRep Q ). k 0 k 0 Gk l W ∈ Q • For V WRepGk l we denote by W V the weight filtration of V and by Gri (V ):= WiV/Wi−1V (i ∈ Z) the associated graded of weight i. As any morphism in WRep Q is strictly compatible with the weight filtration ([J], Lemma 6.8.1), Gk l the functor GrW :WRep Q −→ Rep Q ,V→ GrW (V ) i Gk l Gk l i is exact ([D1], Proposition 1.1.11, ii)) and thus induces a homomorphism

GrW : K (WRep Q ) −→ K (Rep Q ). i 0 Gk l 0 Gk l So we have a homomorphism of abelian groups (5) Φ : K (WRep Q ) −→ K (Rep Q )[T,T−1] ,x→ GrW (x)T i 0 Gk l 0 Gk l i i∈Z

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which is a ring homomorphism as follows from GrW (V ⊗ W ) GrW (V ) ⊗ GrW (W )fori ∈ Z and V,W ∈ WRep Q . i a b Gk l a+b=i W ∗ As Gri (Hc (X)) = 0 for i<0 (see [Ka]) we obtain, composing (4) and (5), a motivic measure −→ Q (6) µk : K0(Vark) K0(RepGk l)[T ] given explicitly by ⎛ ⎞ ⎝ − j W j ⎠ i (7) µk([X]) = ( 1) [Gri (Hc (X))] T . i≥0 j

The slight abuse of notation in denoting (1) and (6) by the same symbol µk will cause no confusion. For a ﬁnite extension k ⊂ L ⊂ k we extend the morphisms induced on K0 by Q Q (2) (resp. (3)) to K0(RepGk l)[T ](resp. K0(RepGL l)[T ]) by demanding that T → T .Withµk as in (6) we then have a commutative diagram

µk / (8) K (Var ) K0(Rep Ql)[T ] 0 Ok GkO

Gk Gk −×kL /k Res Ind GL GL µL / Q K0(VarL) K0(RepGL l)[T ] The commutativity of (8) follows from the compatibility of the weight filtration with restriction and induction, which we leave to the reader to verify using the uniqueness of the weight filtration ([J], Lemma 6.8.1, a)). We now use the weight filtration to establish the following criterion for algebraic independence in K0(Vark). For X/k separated and of finite type and i ≥ 0wewrite W − j W j ∈ Q Gri (X):= ( 1) [Gri (Hc (X))] K0(Rep Gk l) . j W i ∈ Q Then µk([X]) = i Gri (X)T K0(Rep Gk l)[T ]; see (7).

Theorem 2. Let k be ﬁnitely generated and let X1,...,Xn/k be separated and of W ∈ Z ⊆ Q ﬁnite type and assume that Gr0 (Xi) K0(Rep Gk l) for all i and that the W Q Gr1 (Xi) are algebraically independent in K0(Rep Gk l). Then the µk([Xi]) are Q algebraically independent in K0(Rep Gk l)[T ] and hence the [Xi] are algebraically independent in K0(Vark). Q Z Z ⊆ Note that K0(Rep Gk l) is augmented over by the degree map, hence Q K0(Rep Gk l). The last assertion of the Theorem follows from the existence of the motivic measure (6).

Corollary 3. Let k be ﬁnitely generated and let X1,...,Xn/k be proper and smooth. 1 Q Then, if the [Hc (Xi)] are algebraically independent in K0(Rep Gk l),soarethe Q µk([Xi]) in K0(Rep Gk l)[T ] and hence also the [Xi] in K0(Vark). W 0 Proof. As the Xi/k are proper and smooth we have Gr0 (Xi)=[Hc (Xi)] = ⊕| | Q π0(Xi) | |∈Z W − 1 1 [ l ]= π0(Xi) and Gr1 (Xi)= [Hc (Xi)]. Since the [Hc (Xi)] are W algebraically independent by assumption, so are the Gr1 (Xi), and Theorem 2 ap- plies.

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For the proof of Theorem 2 we need some elementary preparation. Let R be ∂ a commutative unitary ring. On R[T1,...,Tn] we have the usual derivations , ∂Ti | | ∈ Nn ∂ α and for α =(α1,...,αn) 0 we write α1 αn for their iterations; here ∂T1 ...∂Tn |α| := αi. Now let f1,...,fn ∈ R[T ],G ∈ Z[T1,...,Tn] be given and put G˜(T ) := G(f1,...,fn) ∈ R[T ]. For N ≥ 1letIN ⊆ R[T ] be the ideal generated by ∂|α|G | |≤ α1 αn (f1,...,fn)for α N and put I0 := 0. The (IN )N≥0 form an ascend- ∂T1 ...∂Tn d ⊂ ≥ ing chain of ideals of R[T ]and dT (IN ) IN+1 for N 0. Proposition 4. For N ≥ 1 we have N ˜ N d G ∂ G df i1 df iN ≡ − N (f1,...,fn) ... (mod IN 1). dT ∂Ti1 ...∂TiN dT dT 1≤i1,...,iN ≤n Proof. We use induction on N ≥ 1. We have ˜ dG d ∂G df i1 = G(f1,...,fn)= (f1,...,fn) , dT dT ∂Ti1 dT 1≤i1≤n i.e. the result for N = 1. We assume N ≥ 2 and compute, using the induction hypothesis: ⎛ ⎞ dN G˜ d ∂N−1G df df ⎝ i1 ··· iN−1 ⎠ N = (f1,...,fn) + α , dT dT ∂Ti1 ...∂TiN−1 dT dT 1≤i1,...,iN−1≤n d some α ∈ IN−2.As (IN−2) ⊆ IN−1 this equals, modulo IN−1: ⎡dT⎛ ⎞ n N df ⎣⎝ ∂ G df j ⎠ df i1 iN−1 (f1,...,fn) ··· ∂Ti1 ...∂TiN−1 ∂Tj dT dT dT 1≤i ,...,i − ≤n j=1 1 N 1 ∂N−1G d + (f1,...,fn) (...) ∂Ti1 ...∂TiN−1 dT N iN :=j ∂ G df i1 df iN ≡ (f1,...,fn) ··· , ∂Ti1 ...∂TiN dT dT 1≤i1,...,iN ≤n as claimed. Proof of Theorem 2. Put Q W ν ∈ R := K0(Rep Gk l),fi(T ):=µk([Xi])= Grν (Xi)T R[T ], ν≥0 W ∈ Z ⊂ ∈ Z ai := Gr0 (Xi) R by assumption. Assume that we have G [T1,...,Tn] with G(µk([X1]),...,µk([Xn])) = 0, i.e.

(9) G˜(T ):=G(f1(T ),...,fn(T )) = 0 . We have to show that G = 0 which follows from the ≥ ∈ Nn | |≤ Claim. For N 0andanyα 0 with α N we have ∂|α|G α1 αn (a1,...,an)=0, ∂T1 ...∂Tn

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which in turn will be established by induction on N.Notethatfi(0) = ai,sofrom (9) putting T =0wegetG(a1,...,an) = 0, i.e. the claim for N = 0. Now assume that N ≥ 1. By applying Proposition 4 to (9) and putting T = 0 we obtain the following relation in R: N ˜ N d G ∂ G df i1 df iN 0= N (T =0)= (a1,...,an) (0) ... (0)+α(0) . dT ∂Ti1 ...∂TiN dT dT 1≤i1,...,iN ≤n

∈ df i W Here α IN−1, hence α(0) = 0 by induction hypothesis. As dT (0) = Gr1 (Xi)we have the following relation in R: N ∂ G W W (10) 0 = (a1,...,an)Gr1 (Xi1 ) ...Gr1 (XiN ) . ∂Ti1 ...∂TiN 1≤i1,...,iN ≤n We collect terms in this expression: Consider { ≤ ≤ }−→{ ∈ Nn | | } π : I := (i1,...,iN ):1 ij n α =(α1,...,αn) 0 : α = N

(i1,...,iN ) −→ (αk := |{1 ≤ j ≤ N : ij = k}|)k=1,...,n .

Given (i1,...,iN ) ∈ I with π((i1,...,iN )) = (α1,...,αn)wehave ∂N G ∂N G = α1 αn ∂Ti1 ...∂TiN ∂T1 ...∂Tn and

W W W α1 W αn Gr1 (Xi1 ) ...Gr1 (XiN )=Gr1 (X1) ...Gr1 (Xn) . So (10) may be written as

N | −1 | ∂ G (11) π (α) α1 αn (a1,...,an) Nn ∂T1 ...∂Tn ∈ 0 α=(α1...αn) |α| =N

· W α1 W αn Gr1 (X1) ...Gr1 (Xn) =0. W ∈ The Gr1 (Xi) R are algebraically independent by assumption and π is surjective, hence |π−1(α)| = 0. So (11) implies ∂N G α1 αn (a1,...,an)=0 ∂T1 ...∂Tn ∈ Nn | | for all α 0 with α = N, concluding the proof of the Claim and of Theorem 2. 2.3. Finite base field. In the situation of Section 2.1 we now assume in addition F that the base field k = q is a finite field. We will use the fact that GFq is topolog- ∈ ically cyclic, generated by the geometric Frobenius Fq GFq , to rewrite the results of Sections 2.1 and 2.2 in terms of characteristic polynomials of Frobenius. We need

to recall some facts about the “universal ring” from [DG], V, 5, no. 2. For any commutative unitary ring R the set Λ(R):=1+tR [[t]] of formal power series with coeﬃcients in R and constant coeﬃcient 1 is an abelian group under multiplication. The resulting functor R → Λ(R) can be endowed with the structure of a functor in commutative unitary rings such that for all R and a, b ∈ R (12) (1 − at)(1 − bt)=1− (ab)t in Λ(R)=1+tR [[t]] ;

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see loc. cit. Section 2.2. The multiplicative unit of Λ(R)is1− t. The ring structure of Λ(R) may cause confusion: The addition is the usual multiplication of power series, but the multiplication in Λ(R) is a rather unusual operation on power series. We will denote addition and multiplication in Λ(R) by the usual symbols but make clear that the composition is to be understood in Λ(R). In this sense, the additive analogue of (12) is (1 − at)+(1− bt)=1− (a + b)t + abt2 in Λ(R). ∈ ∈ ≥ Let R and f Λ(R) be given. One defines ∂ν (f) R for all ν 1bythe − ν−1 df / d t expansion dlog(f)=: ν≥1 ∂ν (f)t . Here, dlog(f)= f (usual quotient of power series) is the logarithmic derivative of f. IfwedenotebyO the identity functor on the category of commutative unitary rings, then each ∂ν :Λ−→ O is a natural transformation (loc. cit. 2.4) and if R is a Q-algebra ∼ N (13) Λ(R) −→ R ,f→ (∂ν (f))ν≥1 is an isomorphism of rings in which RN is the usual product of countably many copies of the ring R. For any ν ≥ 1 there is a natural transformation of ring-valued functors (“Frobenius”) Fν :Λ−→ Λ characterized by ν (14) Fν (1 − at)=1− a t in Λ(R)=1+tR [[t]] , for all R and a ∈ R and a natural transformation of functors in abelian groups ν (“Verschiebung”) Vν :Λ−→ Λ characterized by Vν (1 − at)=1− at in Λ(R). In fact, we have ν (15) Vν (f(t)) = f(t ) for all f(t) ∈ Λ(R)=1+tR [[t]] ; see loc. cit. 2.5. The rings Λ(R) are convenient tools to keep track of characteristic polynomials. Proposition 5. Let k beafieldandV,W finite-dimensional k-vector spaces.

i) For φ ∈ End k(V ) and ψ ∈ End k(W ) we have det(1 − (φ ⊕ ψ)t|V ⊕ W )=det(1− φt|V ) + det(1 − ψt|W ) and det(1−(φ⊗ψ)t|V ⊗W ) = det(1−φt|V )det(1−ψt|W ) in Λ(k)=1+tk [[t]]. For any ν ≥ 1: ν det(1 − φ t|V )=Fν (det(1 − φt|V )) in Λ(k) and ν ∂ν (det(1 − φt|V )) = tr(φ |V ) in k. ii) Let G be a group and N ⊂ G a normal subgroup such that G/N is finite cyclic of order ν generated by the class of π ∈ G.IfV is a k[N]-module, finite-dimensional as a k-vector space, then we have − | G − ν | (16) det(1 πt IndN (V )) = Vν (det(1 π t V )) in Λ(k)=1+tk [[t]] . G ⊗ In ii) we denoted by IndN (V ):=k[G] k[N] V the k[G]-module induced by V . Note that πν ∈ N, hence the right-hand side of (16) is meaningful. Proof. Part i) follows from equations (12) and (14) and [H], Appendix C, Lemma − | G − ν ν | 4.1. Using (15), part ii) is equivalent to det(1 πt IndN (V )) = det(1 π t V ). This has to be checked by an explicit computation which we omit. ForaringR we define a subset R(R) ⊂ Λ(R)by R(R):={f − g | f,g ∈ 1+tR[t]}⊂Λ(R)=1+tR [[t]] . It is clear that R⊂Λ is a subfunctor in abelian groups. One should think of the elements of R(R) as the the power series expansions of certain rational functions, namely the quotients of polynomials with constant coefficient 1.

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Proposition 6. R⊂Λ is a subfunctor in rings.

Proof. First, 1Λ(R) =1− t ∈R(R) ⊂ Λ(R)=1+tR [[t]]. Now it suﬃces to check that for f,g ∈ 1+tR[t]wehavefg ∈R(R) ⊂ Λ(R). Indeed, fg is again a polynomial according to [DG], p. 632, formula (**).

Using Proposition 5 one sees that there is a unique ring homomorphism χ : K (Rep Q ) → Λ(Q ) q 0 GFq l l such that χ ([V ]) = det(1 − F t|V ) ∈ Λ(Q )=1+tQ [[t]] for V ∈ Rep Q q q l l GFq l

and χq is injective as follows from [Alg`ebre], 12,1, Proposition 3. Composing (6) with χq we obtain a motivic measure −→ Q (17) µq : K0(VarFq ) Λ( l)[T ] given explicitly by ⎛ ⎞ ⎝ − j − | W j ⎠ i (18) µq([X]) = ( 1) det(1 Fqt Gri (Hc (X))) T . i≥0 j

Proposition 7. The image of (17) is contained in R(Q)[T ] ⊂ Λ(Ql)[T ].

Proof. Let X/Fq be separated and of ﬁnite type. The cohomological formula for the F Q Q zeta function ZX/Fq (t)ofX/ q ([Mi2], Theorem 13.1) gives in Λ( l)=1+t l [[t]]: − j+1 − | j ZX/Fq (t)= ( 1) det(1 Fqt Hc (X)) ≥ j 0 − j+1 − | W j = ( 1) det(1 Fqt Gri (Hc (X))) ≥ j 0 ⎛ i ⎞ ⎝ − j+1 − | W j ⎠ = ( 1) det(1 Fqt Gri (Hc (X))) . i j≥0 Q ∈ Q In this expression, every summand is in l(t),andwehavethatZX/Fq (t) [[t]]

from the very definition of ZX/Fq (t). For weight reasons there can be no cancellation among different summands (recall that the sums are actually products of power series), hence every summand lies in Ql(t) ∩ Q [[t]] ⊂ Q(t). As every summand is in addition a quotient of polynomials with constant coefficient 1, every summand lies in fact in R(Q) and comparison with (18) concludes the proof.

The relevance of having a motivic measure taking values in R(Q)[T ] rather than in Λ(Ql)[T ] will become clear in Section 3.2. Finally we record some formulae relating the various operations on Λ with the motivic measure µq. Proposition 8. i) If X/F is separated and of ﬁnite type, then, for all ν ≥ 1, q − j ν | W j i Q ∂ν (µq([X])) = i≥0 j( 1) tr(Fq Gri (Hc (X))) T in [T ].

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ii) For Fq ⊂ Fqν ⊂ Fq the ﬁnite extension of degree ν ≥ 1 we have a commutative diagram

µq / K (VarF ) R Q (19) 0 O q ( )[OT ] −× F Fq qν /Fq Fν Vν µqν /R Q K0(VarFqν ) ( )[T ]

In i) ∂ν denotes the restriction of ∂ν :Λ(Q) −→ Q to R(Q) ⊂ Λ(Q) extended to ∂ν : R(Q)[T ] −→ Q[T ] by demanding that ∂ν (T )=T , and in ii) Fν and Vν denote the maps which on R(Q) are the restrictions of Fν ,Vν :Λ(Q) −→ Λ(Q)andwith Fν (T )=T and Vν (T )=T . Proof. Part i) follows from (18) and Proposition 5i). Part ii) follows from (8) and Proposition 5i) and ii).

3. Algebraic independence 3.1. Virtual continuous representations. In view of Corollary 3 we are led to consider the following problem: Given a topological group T and a finite extension K/Ql denote by Rep T K the category of finite-dimensional continuous representations of T over K.Given finitely many Vi ∈ Rep T K, when are their classes [Vi] ∈ K0(Rep T K) algebraically independent? In this subsection we will use a Tannakian argument to reduce this question to a problem about algebraic independence of representations of a (possibly non- connected) reductive algebraic group and in a special case of a diagonalizable group. For this latter case we establish a rather explicit Jacobi criterion (Lemma 10). The subtleties arising from the possible non-connectedness of the reductive group above will be illustrated by an example (Theorem 11) and motivate the following definition.

Deﬁnition 9. Representations Vi ∈ Rep T K are geometrically algebraically independent if for all open subgroups of ﬁnite index T ⊂ T the classes of the restrictions T ∈ [ResT (Vi)] K0(Rep T K) are algebraically independent.

We now give a series of reduction steps to obtain workable criteria for the [Vi] ∈ K0(Rep T K) to be algebraically (in)dependent. The reader is invited to glance at a simple application, Theorem 11, ﬁrst. ss Denoting by Vi the semi-simpliﬁcation of Vi we have ss (20) [Vi]=[Vi ]inK0(Rep T K),

and we can assume the Vi to be semi-simple themselves. In the sequel we will repeatedly use the obvious fact that elements xi ∈ R ⊂ S of a subring R of S are algebraically independent in R if and only if they are so in S.Wenowuse a Tannakian argument to find an accessible subring of K0(Rep T K) containing all the [Vi]. The reference for Tannakian categories is [D2]. With the obvious notion of tensor product, dual and using the forgetful functor Rep T K −→ VectK to the category of finite-dimensional K-vector spaces as a fibre functor, the category n Rep T K is a neutral Tannakian category over K. We put V := i=1 Vi and consider ⊗ i : C := V → Rep T K,

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the Tannakian subcategory generated by V , i.e. the smallest strictly full subcategory of Rep T K containing V and stable under the formation of ﬁnite direct sums, tensor products, duals, subobjects and quotients. As i is an exact tensor functor it induces a ring homomorphism

(21) K0(C) → K0(Rep T K) which is injective: one checks that i preserves simple objects and uses the fact that the K0 of an abelian category with all objects of ﬁnite length is Z-free on the

isomorphism classes of its simple objects [Q], 5, Corollary 1. By construction we have [Vi] ∈ K0(C), and our initial problem is reduced to considering the algebraic (in)dependence of the [Vi]aselementsofK0(C). Let (22) φ : T −→ Gl (V )(K) be the continuous homomorphism giving the action of T on V .HereGl(V ) denotes the algebraic group and Gl (V )(K) its group of K-points endowed with the topology inherited from the one of K. The Zariski closure of the image of φ,say (23) G := im(φ) ⊂ Gl (V ), is an affine algebraic group over K ([Bo], Proposition 1.3, b)) and can be identified with the Tannaka group of C as follows ([S4], 1.3). From φ we have a continuous homomorphism φ˜ : T −→ G(K), and the restriction of the fibre functor C−→ VectK induces an equivalence of Tannakian categories ∼ (24) C −→ Rep G, K where Rep G denotes the category of finite-dimensional representations of the K algebraic group G over K. A quasi-inverse of (24) is given by ∼ ψ Rep G −→ C , (G → Gl (W )) → ψ(K) ◦ φ.˜ K As C is semi-simple by (20), then [DMOS], Remark 2.28, implies that G is (not necessarily connected) reductive. The isomorphism K (C) K (Rep G) induced 0 0 K by (24) reduces our initial problem to studying the algebraic (in)dependence of given representations of a reductive group. We next show that we may allow finite extensions of the coefficients K of these representations. For K ⊂ L a finite extension, the exact tensor functor Rep G −→ Rep G ⊗ L, V → V ⊗ L K L K induces a ring homomorphism (25) K (Rep G) → K (Rep G ⊗ L) 0 K 0 L which is injective; cf. [Mi1], Lemma 3.14, for a similar argument. This allows us to replace G by G ⊗ L and to assume that the connected component of identity G0 ⊂ G is a connected split reductive group and that the finite étale group G/G0 is in fact constant. Extending K further to contain a splitting field of G/G0, the ring K (Rep G/G0) is a familiar object from the representation theory of 0 K finite (abstract) groups [S3], Chapter II. Also K (Rep G0) can be described quite 0 K precisely (cf. below) but without imposing further, rather restrictive, conditions (e.g. G G0 × G/G0) I cannot say much about the structure of K (Rep G). 0 K We will describe two special situations in which the above line of thought may be continued.

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So K/Ql is a finite extension, G is a split reductive group over K and we need to study algebraic (in)dependence in the ring K (Rep G). We assume in addition 0 K that G is connected and fix a K-split maximal torus S ⊂ G. Then restriction induces an injective ring homomorphism K (Rep G) → K (Rep S) 0 K 0 K [S1], 3.4, and the structure of K (Rep S) is particularly simple; cf. (28) for 0 K the more general case of a diagonalizable group. It then becomes important to determine the character group of S which in the present case (the topological group T being arbitrary) we do not know how to do. The most satisfactory solution of our initial problem can be obtained in the case that T = F is topologically cyclic. Though very special, this case will be of interest for us because it covers the case of the Galois group of a finite field. So let

φ : T = F −→Gl (V )(K) Gl m(K) n be as in (22) (so m = i=1 dim K Vi). By our preliminary reductions (20) and (25) and conjugating φ(F ) suitably we can assume that

(26) φ(F )=diag(α1,...,αm) ∈ Dm(K) ∗ for some αi ∈ K ,whereDm ⊂ Gl m is the torus of diagonal matrices. Then G as in (23) is the smallest algebraic subgroup G ⊂ Gl m defined over K such that φ(F ) ∈ G(K), and D := G ⊂ Dm is a closed subgroup, hence diagonalizable by [Bo], Proposition 8.4. The character group X(D)ofD can be identified with the subgroup of K∗ generated by the eigenvalues of φ(F ) as follows from [Bo], 8.2, Corollary: ∗ (27) X(D) α1,...,αm⊂K . We recall the structure of K (Rep D) from [S1], 3.4. Giving V ∈ Rep D is 0 K K equivalent to giving a “grading of type X(D)” of the finite-dimensional K-vector ⊂ space V , i.e. V = χ∈X(D) Vχ,whereVχ V is the χ-eigenspace and ∼ (28) ch : K (Rep D) −→ Z[X(D)] , [V ] → dim (V )eχ 0 K K χ χ is a ring isomorphism. Here Z[X(D)] is the group algebra over X(D), i.e. Z[X(D)] is Z-free on the set {eχ : χ ∈ X(D)} and eχeχ = eχ+χ . Decomposing the finitely generated abelian group X(D)as d tor X(D)=X(D) Zei, i=1 tor where X(D) ⊂ X(D) is the torsion subgroup and the ei ∈ X(D) are suitably chosen, the natural map d tor ∼ t f t+f Z[X(D) ] ⊗Z Z Zei −→ Z[X(D)] ,e⊗ e → e i=1 is an isomorphism and d ∼ − − Z Z −→ Z 1 1 aiei → a1 ad ei [T1,...,Td,T1 ,...,Td ] ,e T1 ...Td i=1

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is an isomorphism to the ring of Laurant polynomials over Z in d=rk(X(D)) = dim(D) variables. Since X(D)tor is finite, A := Z[X(D)tor] is a finite flat Z-algebra and summing up we have (29) K (Rep D) A[T ,...,T ,T−1,...,T−1]. 0 K 1 d 1 d Note that, if X(D) is as in (27), then X(D)tor ⊂ K∗ is a finite subgroup, hence cyclic of order N, say. Then we have (30) A = Z[Z/N ] Z[S]/(SN − 1). So our initial problem is finally reduced to deciding whether the ∈ −1 −1 fi := ch([Vi]) A[T1,...,Td,T1 ,...,Td ],

i.e. the characters of the given representations Vi, are algebraically independent. For this problem one can prove a Jacobian criterion as follows. Let A be a finite flat Z-algebra. We will prove a criterion for given ∈ −1 −1 f1,...,fn A[T1,...,Tm,T1 ,...,Tm ] to be algebraically independent in terms of the Jacobian ∂f J := i ∂Tj 1≤i≤n, 1≤j≤m −1 −1 which is an n by m matrix with entries from A[T1,...,Tm,T1 ,...,Tm ]. For a homomorphism φ : A → Q we denote by ⊗ −1 −1 −1 ⊗ Z −1 φ 1:A[T ,T ]:=A[T1,...,Tm,T1 ,...,Tm ]=A [T ,T ] → Q −1 Q −1 −1 Q ⊗ Z −1 [T ,T ]:= [T1,...,Tm,T1 ,...,Tm ]= [T ,T ] the base change of φ.Then(φ ⊗ 1)(J) is a matrix with coefficients in −1 Q[T ,T ] ⊂ Q := Q(T1,...,Tm), where Q is the field of rational functions in m variables over Q. For a homomorphism ψ : Q[T ,T−1] −→ K with K some field, (ψ ◦ (φ ⊗ 1))(J)isamatrix with coefficients in K. Finally, by rkL(M) we will denote the rank of a matrix M having coefficients in the field L. The Jacobi criterion for algebraic independence mentioned above is the following. Lemma 10. Keeping the above notations we have: 1) The following are equivalent: −1 i) The fi ∈ A[T ,T ] are algebraically dependent. ii) For all φ : A → Q we have

rkQ((φ ⊗ 1)(J))

rkK (ψ ◦ (φ ⊗ 1)(J)) = n.

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The proof of Lemma 10 consists in a reduction, which we will omit, to the case when A is a ﬁeld of characteristic zero. In this case the Lemma is [Bo], Chapter AG, Theorem 17.3. We conclude this subsection by giving two applications.

Theorem 11. Let k be a ﬁnite ﬁeld and let E1,E2/k be non-isogeneous ordinary elliptic curves. Let E2/k be the quadratic twist of E2, and consider the abelian × × surfaces A1 := E1 E2 and A2 := E1 E2 over k and the associated Galois 1 ∈ Q ⊂ representations Vi := Hc (Ai) Rep Gk l.Letk L be the unique quadratic extension of k inside k.Then: Gk Gk i) ResG (V1) ResG (V2). L L Q ii) The classes [V1] and [V2] are algebraically independent in K0(Rep Gk l). In particular, the classes [A1] and [A2] are algebraically independent in K0(Vark).

Remark. The representations V1 and V2 are algebraically independent but not geometrically algebraically independent. In fact, after restriction to the subgroup of index two of Gk their classes become algebraically dependent and in fact equal. We also see that x := [A1] − [A2] ∈ K0(Vark) generates a polynomial ring over Z inside K0(Vark) even though x lies in the kernel of the base change homomorphism K0(Vark) −→ K0(VarL). This shows that K0(Vark) encodes ﬁne arithmetic invariants of varieties over k.

Proof. Part i) is clear because A1 ×k L A2 ×k L, and the last assertion follows ∗ from part ii) and Corollary 3. We prove ii): Fix πi ∈ Q a Weil number attached to Ei/k and denote by πi the conjugate of πi.NotethatQ(π1)andQ(π2)aredistinct imaginary quadratic ﬁelds, hence

(31) Q(π1) ∩ Q(π2)=Q.

1 The eigenvalues of the geometric Frobenius Fk acting on the Hc of E1,E2 and { } { } {− − } E2 are (respectively) π1, π1 , π2, π2 and π2, π2 . Fixing a ﬁnite extension Ql ⊂ K containing π1 and π2 we get in the notation of (26):

φ(Fk)=diag(π1, π1,π2, π2,π1, π1, −π2, −π2) ∈ D8(K) and ∗ X(D) π1, π1,π2, π2, −1⊂K .

Using (31) and the fact that E1 and E2 are ordinary, one checks that the structure of X(D)isasfollows: {± }⊕ Z ⊕ Z ⊕ Z (32) X(D)= 1 π1 (π1) π2 . From (32) we get as in (29) together with (30)

(33) K (Rep D) (Z[S]/(S2 − 1))[T ,T−1,...,T ,T−1] 0 K 1 1 3 3

π π1 π π2 and compute the characters fi := ch([Vi]) as f1 = e 1 + e + e 2 + e and − π1 π1 −π2 −π2 1 f2 = e + e + e + e .Wehaveπ2 = π1π1π2 and using this, (32) and (33) we get (denoting abusively the images of [Vi] under (33) by fi again)

−1 f1 = T1 + T2 + T3 + T1T2T3 and −1 f2 = T1 + T2 + ST3 + ST1T2T3 .

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To show that f1 and f2 are algebraically independent we compute the Jacobian of f and f 1 2 −1 −1 − −2 1+T2T3 1+T1T3 1 T1T2T3 J = −1 −1 − −2 1+ST2T3 1+ST1T3 S ST1T2T3 which we specialize by putting S = −1, T = −1, T = T = 1 to obtain 1 2 3 20 2 02−2

which has rank two, hence f1 and f2 are algebraically independent by Lemma 10, as was to be shown.

Our next application of the methods developed so far is the following. Theorem 12. Let k be a ﬁnite ﬁeld. Then there is a sequence of proper, smooth and geometrically connected curves Xi/k (i ≥ 1) such that the classes [Xi] ∈ K0(Vark) are algebraically independent.

Theorem 13. Let k be a number field and {Ei}i∈I a set of elliptic curves over k Z such that the Ei are pairwise non-isogeneous and satisfy End k(Ei)= . Then the classes [Ei] ∈ K0(Vark) are algebraically independent. Note that over any number field there are infinite sets of elliptic curves satisfying the assumptions of Theorem 13. To prove these results we will need the following sufficient criterion for geometric algebraic independence in terms of Frobenius eigenvalues.

Lemma 14. Let Zˆ denote the pro-ﬁnite completion of Z and let K/Ql be a ﬁnite extension and let V1,...,Vn ∈ Rep ZˆK be given. For 1 ≤ i ≤ n put mi := dim K Vi and let { } ⊂ ∗ λi,j 1≤j≤mi K

be the eigenvalues of F := 1 acting on Vi. For any 1 ≤ k ≤ n consider the ﬁnite- dimensional Q-vector space ⊗ Q (34) Ak := λi,j 1≤i≤k, 1≤j≤mi Z

and set A0 := 0.If

(35) dim QAk > dim QAk−1 for k =1,...,n,

then the V1,...,Vn ∈ Rep ZˆK are geometrically algebraically independent. For the application of this lemma in the proof of Theorem 13 we make its assumptions more explicit in the case when the Vi are (dual to) the Tate-modules of elliptic curves. F Zˆ Lemma 15. Let q be a finite field and identify GFq = using the geometric Frobenius. Let E1,...,En/Fq be elliptic curves, assume that E1 is ordinary, put V := H1(E ) ∈ Rep (Q ) and define the A as in Lemma 14. Then the assump- i c i GFq l i tions of this Lemma are satisfied if and only if dimQ An = n +1.

Proof. Since E1 is ordinary we have dimQ A1 =2.Fork =2,...,n we have dimQ Ak ≤ dimQ Ak−1 + 1 because λk,1λk,2 = q ∈ Ak−1. The assertion is obvious now.

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Proof of Lemma 14. In the notation of (26) we have φ(F )=diag(λ ,...,λ ,...,λ ,...,λ ) ∈ D n (K). 1,1 1,m1 n,1 n,mn i=1 mi

As a ﬁrst step we will prove that the characters fi := ch([Vi]) ∈ K0(Rep D)are K algebraically independent. Here D ⊂ D n is the smallest algebraic subgroup i=1 mi with φ(F ) ∈ D(K). After suitably enlarging K (legitimate by (25)) we identify (as in (27)) the character group X(D)ofD, ⊂ ∗ (36) X(D) λi,j 1≤i≤n, 1≤j≤mi K , and choose a decomposition N Z (37) X(D)= ζ πi i=1 ∗ ∗ with ζ ∈ K a root of unity of order M,say,andπ1,...,πN ∈ K a Z-basis of the free part of X(D). We order the πi’s such that their images πi ⊗1inX(D)⊗Q = An ≤ ≤ ⊗ ⊗ satisfy the following: For all 1 k n, π1 1,...,πdk 1isabasisofAk.Here dk := dimQ Ak and (35) gives

(38) d0 =0

(39) N =dim(D)=dimQ(X(D) ⊗ Q)) (36) ⊗ Q =dimQ( λi,j 1≤i≤n, 1≤j≤mi ) (34) (35) =dimQ(An) ≥ n.

For any 1 ≤ i ≤ n and 1 ≤ j ≤ mi we decompose, according to (37),

ai,j a1,i,j aN,i,j (40) λi,j = ζ π1 ...πN

for suitable ai,j ,ak,i,j ∈ Z.Using(29) K (Rep D) A[T ,T−1,...,T ,T−1] 0 K 1 1 N N where, according to (30), we have A Z[S]/(SM − 1). The characters under consideration are given as follows (taking (33) as an identiﬁcation):

mi mi (40) ai,j a1,i,j aN,i,j (41) fi =ch([Vi]) = λi,j = ζ π1 ...πN j=1 j=1 mi a a ai,j 1,i,j N,i,j ≤ ≤ = S T1 ...TN , 1 i n. j=1 We deﬁne φ : A −→ Q by S → 1and Q −1 −1 → Q ψ : [T1,T1 ,...,TN ,TN ] Q := (T1,...,TN )

to be the natural inclusion. Now we show that the [Vi] ∈ K0(Rep ZˆK) are algebraically independent. Using Lemma 10 it suﬃces to show that ψ ◦ (φ ⊗ 1)(J)has ∂fi ≤ rank n over Q.HereJ is the Jacobi matrix J = ∂T .Sincen N, j 1≤i≤n, 1≤j≤N by (39) this is equivalent to the vectors ∂fi vi := ψ ◦ (φ ⊗ 1) , 1 ≤ i ≤ n, ∂Tj 1≤j≤N

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1668 N. NAUMANN ∂fi being linearly independent over Q. Let us abbreviate xi,j := ψ ◦(φ ⊗ 1) ∈ Q, ∂Tj hence vi =(xi,j )1≤j≤N . We ﬁrst prove the following claim concerning the shape of the matrix (xi,j ).

Claim. For every 1 ≤ i ≤ n there is some 1 ≤ ji ≤ N such that (42) xi ,ji =0fori

xi,ji =0.

Indeed, given 1 ≤ i ≤ n let ji be any integer satisfying di−1

0inQ which by (42) simplifies to αkxk,jk =0withxk,jk = 0, hence αk =0,a contradiction. At this point the proof of the fact that the classes [Vi] ∈ K0(Rep ZˆK) are algebraically independent is complete. To show that they are in fact geometrically algebraically independent, let an open subgroup of finite index T ⊂ Zˆ be given. There is an integer L ≥ 1 such that T = LZˆ and the set of eigenvalues of the topological generator L ∈ T acting on Zˆ { } { L } ResT (Vi)is λi,j i,j = λi,j i,j . So the homomorphism −→ → L λi,j 1≤i≤k,1≤j≤mi λi,j 1≤i≤k,1≤j≤mi ,x x , is surjective with finite kernel and induces an isomorphism −→∼ ⊗ Q Ak λi,j 1≤i≤k,1≤j≤mi Z ≤ ≤ Zˆ for all 1 k n. Thus the tuple (T , ResT (Vi)) satisfies the analogue of con- Zˆ ∈ dition (35), and by what has already been proved we see that the [ResT (Vi)] K0(Rep T K) are algebraically independent. As T was arbitrary this shows that the Vi are in fact geometrically algebraically independent and concludes the proof. Before engaging into the slightly involved proof of Theorem 12 we remark that if in this result one replaces curves by abelian varieties one can give a quick proof ∗ as follows: The subgroup Γ ⊂ Q generated by q-Weil numbers of weight one has infinite rank. Choosing a suitable sequence of such Weil numbers and corresponding abelian varieties (which exist by T. Honda), one obtains the result using Lemma 14 and Corollary 3. Even though rk(Γ) = ∞ is well known we could not find it stated explicitly in the literature, so we record this result (which will be clear from our proof of Theorem 12) for reference. ∗ Corollary 16. Let q be a prime power. Then the subgroup of Q generated by q-Weil numbers of weight one has infinite rank. We now start the proof of Theorem 12. In fact, we will give a family of curves satisfying the conclusion of Theorem 12 explicitly.

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Let Fp ⊂ k be the prime field of k and for any a ≥ 1letYa/Fp be the proper smooth curve which is birational to the affine plane curve given by a yp − y = xp −1. 1 a − − The Ya are geometrically connected of genus 2 (p 2)(p 1). The arithmetic of these curves has been studied by H. Davenport and H. Hasse [DH]. Unfortunately, we cannot assert that the sequence {Ya}a≥1 satisfies the conclusion of Theorem 12 but a suitable subsequence will do. Let T = T (p) be the integer determined by Lemma 19 below and let {ni}i≥1 be the sequence of prime numbers greater than T ordered increasingly. Then the ⊗ ≥ Xi := Yni Fp k, i 1, are proper, smooth and geometrically connected curves over k, and the following result makes the existence statement of Theorem 12 explicit.

Theorem 17. The classes [Xi] ∈ K0(Vark)(i ≥ 1) are algebraically independent. For any i ≥ 1letuswriteV := H1(Y ) ∈ Rep Q . Using Corollary 3, i c ni GFp l Theorem 17 results from the following. Theorem 18. The classes V ∈ Rep Q (i ≥ 1) are geometrically algebraically i GFp l independent. This will be proved using Lemma 14 for which we need to review some results concerning the Weil numbers attached to the Jacobians of the curves Ya/Fp. We fix a valuation v of Q lying over p and normalized by v(p) = 1. We will denote by the same letter v the restriction of v to any subfield K ⊂ Q. F a Let Fp be the geometric Frobenius of p. Then the eigenvalues of Fp acting on 1 Hc (Y a) are given by generalized Jacobi sums t (a) t 2πi a (43) τj(χ ) := − χ (u)exp · j · tr(u) , 1≤j ≤p − 1, 1≤t≤p − 2, ∈F∗ p u pa F∗ → C∗ F∗ where χ : pa is a certain faithful character of pa (a Teichmüller lift de- F −→ F pending on the choice of v;see[Ma],4) and tr : pa p denotes the trace, [Y], (2). t (a) t (a) a We will abbreviate τ(χ ) := τ1(χ ) in the following. Let 1 ≤ t ≤ p − 2be given and let a−1 i t = jip , 0 ≤ ji ≤ p − 1, i=0 a−1 be the p-adic expansion of t and put σ(t):= i=0 ji.Then[Ma],4, gives the valuations of the generalized Jacobi sums as σ(t) (44) v(τ(χt)(a))= . p − 1 We will need the following lower bound on Euler’s phi-function, φ(n):=|(Z/n)∗| for n ≥ 1. Lemma 19. Let p be a prime number. Then there is an integer T = T (p) such that φ(pn − 1) >φ(p − 1) for all n ≥ T. n

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Proof. This follows from the inequality, valid for any m ≥ 1, m m − 1 ≥ φ(m) ≥ , C log log m with a suitable positive constant C. The ﬁrst inequality is trivial and the second is [RS], Theorem 15.

Proof of Theorem 18. It suffices to show that for any given M ≥ 1theV1,...,VM ∈ Rep Q are geometrically algebraically independent. So we fix some M ≥ 1. For GFp l ≤ ≤ { } ⊂ Q∗ any 1 i M let λi,j 1≤j≤(pni −2)(p−1) be the eigenvalues of Fp acting on Vi.Notethat { ni } { t (ni)} (45) λi,j 1≤j≤(pni −2)(p−1) = τj(χ ) 1≤j≤p−1, 1≤t≤pni −2 ∗ by (43). We define B0 := {1}⊂Q and for 1 ≤ k ≤ M we define Bk := ∗ ∗ λi,j 1≤i≤k,allj ⊂ Q to be the subgroup of Q generated by the Fp-eigenvalues occurring among the V1,...,Vk and Ak := Bk ⊗Z Q ,dk := dim QAk.Using Lemma 14 the proof will be complete if we establish the following:

(46) We have dk >dk−1 for k =1,...,M.

As Bk−1 ⊂ Bk we always have dk ≥ dk−1, and we need to show that this inequality is strict. For any n ≥ 1wedenotebyKn ⊂ Q the ﬁeld of n-th roots of unity and ∗ we ﬁx isomorphisms Gal(Kn/Q) (Z/n) using a compatible system of roots of ∗ unity in Q . Recall that Kn ∩ Km = Kgcd(n,m) and KnKm = Klcm(n,m),wheregcd and lcm mean greatest common divisor and least common multiple, respectively. From (45) and (43) we see that ni ∈ (47) λi,j Kpni −1Kp = Kp(pni −1), k ≤ ≤ hence, introducing Nk := i=1 ni for 0 k M,wehavethat Nk { Nk ∈ }⊂ (48) Bk := x : x Bk Kp(pNk −1) for k =0,...,M. We now prove (46) by contradiction, assuming that we are given some 1 ≤ k ≤ M such that

(49) dk = dk−1.

As k will be ﬁxed from now on, we put n := nk and N := Nk−1 to ease the reading. The cyclotomic ﬁelds to play a role are organized in the following diagram:

(50) Kp(pn−1) XX Kp(pN −1) s XXXXX s HH ss XXXXX ss H ss XXXXsXs HH ss ss XXXXX HH ss ss XXXXX HH s s XXXXX Kpn−1 KpN −1 m Kp KKK s mmm KK sss mmm KKK ss mmm KK sss mmm s mmm mmm Kp−1 mmm mmm mmm mmm mmm mmm Q Observe that by construction gcd(n, N)=1(N is a product of primes diﬀerent from the prime n) and hence gcd(pn − 1,pN − 1) = p − 1 which implies as recalled

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above that Kpn−1 ∩ KpN −1 = Kp−1 as asserted by (50). The remaining assertions implicit in (50) are checked similarly. According to (43) there are λ1 and λ2 which are eigenvalues of Fp acting on Vk (hence λi ∈ Bk by the deﬁnition of Bk)andsuch that n 1 (n) (51) λ1 = τ(χ ) , n pn−2 (n) λ2 = τ(χ ) ; n ∈ hence λi Kp(pn−1) by (43) and by [Y], (14) we have n(p−1) ∈ (52) τi := λi Kpn−1 ,i=1, 2.

The idea now is the following. We will exhibit two places v1 and v2 of Kpn−1 lying above the same place of Kp−1 such that a suitable linear combination x of τ1 and τ2 has diﬀerent valuations at v1 and v2, and using (49) we will see that we may choose x to lie in Kp−1 which will give the desired contradiction. Here are the details. The inclusion Kp−1 ⊂ Kpn−1 corresponds to a surjection of Galois groups π :(Z/pn − 1)∗ //(Z/p − 1)∗.

The decomposition group of the place v1 := v of Kpn−1 is generated by the residue of p (since p is unramiﬁed in Kpn−1)andπ factors over a surjection (53) π :(Z/pn − 1)∗/p //(Z/p − 1)∗. The left-hand side of (53) acts simply transitively by conjugation on the set of places of Kpn−1 lying above p, and this will be used to construct a second such place v2. The order of the left-hand side of (53) is φ(pn − 1)/n > φ(p − 1) = |(Z/p − 1)∗| by Lemma 19 and our choice that n ≥ T (p)sowemaychoose (54) 1 = γ ∈ ker (π). −1 We denote by v2 the place of Kpn−1 conjugate by γ to v1, i.e. v2(x)= ∈ ∗ v1(γ.x)=v(γ.x) for all x Kpn−1. Now we compute some valuations:

(52) (51),(44) σ(1) v (τ ) =(p − 1)v(λn) =(p − 1) =1and 1 1 1 p − 1 n (52) (51),(44) σ(p − 2) v (τ ) =(p − 1)v(λn) =(p − 1) = n(p − 1) − 1, 1 2 2 p − 1 using for the computation of σ(pn − 2) that n−1 n−1 pn − 2=pn − 1 − 1= (p − 1)pi − 1=(p − 2) + (p − 1)pi. i=0 i=1 Let the p-adic expansion of γ be given by n−1 i (55) γ = αip , 0 ≤ αi ≤ p − 1. i=0 Observe that n−1 (54) σ(γ)= αi ≡ π(γ) ≡ 1(mod(p − 1)). i=0

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If we had σ(γ) = 1, then [Y], Lemma 3, ii), would imply that γ ∈ pZ, hence γ =1 in (Z/pn − 1)∗/p, contrary to our choice (54). So we have (56) σ(γ)=1+a(p − 1) for some a ≥ 1.

Now we compute the valuations at v2:

(52),(51) 1 (n) 1 (n) v2(τ1) =(p − 1)v2(τ(χ ) )=(p − 1)v(γ.τ(χ ) ) (44) σ(γ) (56) =(p − 1)v(τ(χγ)(n)) =(p − 1) =1+a(p − 1). p − 1

In the third term we consider γ as an element of Gal(Kpn−1/Q), and the second t (n) equality follows from our deﬁnition of v2. The third equality uses γ.τ(χ ) = τ(χγt)(n) which follows from (43). Similarly we have

(52),(51) pn−2 (n) γ(pn−2) (n) v2(τ2) =(p − 1)v2(τ(χ ) )=(p − 1)v(τ(χ ) ) (44) σ(−γ) (∗) =(p − 1)v(τ(χ−γ)(n)) =(p − 1) = σ(−γ) = n(p − 1) − σ(γ) p − 1 (56) = n(p − 1) − (1 + a(p − 1)) = (n − a)(p − 1) − 1. In this sequence of equalities, we would like to justify (*), which is an elementary property of the function σ: From (64) we get the p-adic expansion of −γ:

n−1 n−1 n−1 n− i (mod p 1) n i i −γ = − αip ≡ p − 1 − αip = (p − 1 − αi)p , i=0 i=0 i=0 hence n−1 n−1 σ(γ)+σ(−γ)= αi + (p − 1 − αi)=n(p − 1), i=0 i=0 as used in (*). To sum up, we have computed the matrix v (τ ) v (τ ) 1 n(p − 1) − 1 X := 1 1 1 2 = v2(τ1) v2(τ2) 1+a(p − 1) (n − a)(p − 1) − 1 and compute its determinant (57) det(X)=(n − a)(p − 1) − 1 − (1 + a(p − 1))(n(p − 1) − 1) =(n − a)(p − 1) − 1 − (n(p − 1) − 1+an(p − 1)2 − a(p − 1)) = −an(p − 1)2 =0 , ∈ Z µ ν ∈ because a = 0 by (56). If µ, ν are given and we put x := τ1 τ2 Kpn−1 we have v (x) µ 1 = X . v (x) ν 2 ≥ ∈ Z µ R By (57) there is some integer R 1andµ, ν such that X (ν )= 0 ,soif µ ν ∈ we put x := τ1 τ2 for these particular values of µ and ν we get x Kpn−1 with

(58) v1(x)=R ≥ 1andv2(x)=0.

Note that for the restrictions of v1 and v2 to Kp−1 we have | | (59) v1 Kp−1 = v2 Kp−1

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because the restriction of γ ∈ Gal(Kpn−1/Q)toKp−1 is trivial, i.e. π(γ)=1;cf. (54). Now we can ﬁnally derive a contradiction. By (49) we know that Bk/Bk−1 is torsion. As λ1,λ2 ∈ Bk we ﬁnd an integer S ≥ 1with

(48) S ∈ N ⊂ (60) λi Bk−1 Kp(pN −1) ,i=1, 2

S µ ν S (52) µ ν n(p−1)S (recall that N = Nk−1), and furthermore x =(τ1 τ2 ) =(λ1 λ2 ) ,so S S x ∈ Kpn−1 by (52) and x ∈ Kp(pN −1) by (60), from which we conclude that

S (50) (61) x ∈ Kpn−1 ∩ Kp(pN −1) = Kp−1.

(58) S (61),(59) S (58) This implies SR = v1(x ) = v2(x ) = 0, a contradiction which concludes the proof.

Proof of Theorem 13. We can assume that the given set of elliptic curves {Ei}i∈I { } 1 = E1,...,En is finite. We will show that the classes of Vi := Hc (Ei)in Q K0(Rep Gk l) are algebraically independent which is sufficient by Corollary 3. We Z ⊂ Q consider the subring R := [[Vi]] K0(Rep Gk l) generated by the [Vi]. For any finite place v of k with residue field k(v)atwhichalltheEi have good reduction, −→ Q we have a canonical homomorphism φv : R K0(Rep Gk(v) l) because all Vi are unramified at v. We are going to construct such a place v so that theφv([Vi]) ×···× 1 n are algebraically independent. For A := E1 En we have Hc (A) i=1 Vi ⊂ 1 Q and denote by Gl Gl (Hc (A))( l) the closure of the image of Galois. Since Z 1 End k(Ei)= we know that the Hodge group Hdg(H (Ei)) is isogeneous to Sl2. 1 As the Ei are pairwise non-isogeneous we have that Hgd(H (A)) is isogeneous to n 1 Sl2 , and hence the Mumford-Tate group MT(H (A)) has rank n +1. Nowobserve that the Mumford-Tate conjecture is known to be true for A/k:Forn = 1 it is a celebrated result of J-P. Serre, and one reduces to this case using [R], Lemma on page 790. We conclude that Gl has rank n+1. By [S5] we find a finite place v of k of ⊂ 0 good reduction for all Ei and such that the corresponding Frobenius torus Tv Gl is maximal, i.e. Tv has rank n + 1. But this rank is nothing other than dimQ An in the notation of Lemma 15 (applied to the reductions at v of the Ei), and using this Lemma we conclude that the φv([Vi]) are indeed algebraically independent as was to be shown.

3.2. Using a result of Skolem. In the previous subsection we used a Tannakian argument to establish the algebraic independence of l-adic Galois representations given by the cohomology of suitable varieties. In this subsection we approach the same goal using a lemma of Skolem on the shape of the power series expansion of rational functions together with the rationality properties given in Section 2.3. The possibility of this was suggested by reading a note of J-P. Serre [S2]. Let k be a ﬁnite ﬁeld. We will use the motivic measure from Proposition 7,

µk : K0(Vark) −→ R (Q)[T ], to give sufficient conditions for the classes of given varieties over k to be algebraically independent in K0(Vark). The main observation is that, if there is a non-trivial algebraic relation among the motivic measures of some varieties, then, at least after a finite extension of the base field, there will be an irreducible relation; see Section 3.3.

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For ν ≥ 1wedenotebykν the unique extension of degree ν of k inside some ﬁxed algebraic closure of k. 3.3. Irreducible equations.

Theorem 20. Let k be a finite field and let X1,...,Xn/k (n ≥ 1) be separated and of finite type and assume that µk([X1]),...,µk([Xn]) are algebraically dependent in R(Q)[T ]. Then there are M ≥ 1 and an irreducible polynomial G ∈ Z[T1,...,Tn] such that × × G(µkM ([X1 k kM ]),...,µkM ([Xn k kM ])) = 0. Remark. This result is non-trivial even for n = 1 because R(Q)[T ] contains zero divisors. In the remark following the proof of Theorem 25 we give an example showing the necessity of allowing the finite extension k ⊂ kM in Theorem 20. The starting point of the proof of Theorem 20 is a result due to Skolem which we now explain. For brevity, we will call a subset X ⊂ N good if it is the union of a finite set and finitely many arithmetic progressions for a single modulus, i.e. X is good if there are Σ ⊂ N finite, M ≥ 1andI ⊂{0,...,M − 1} such that X =Σ (i + MN). i∈I The collection of good subsets is stable under finite union and finite intersection. Skolem’s result is the following [S2].

Proposition 21. Let f ∈ Q(t) be a rational function with power series expansion ν { ≥ } f = ν∈Z aν t . Then the set ν 0:aν =0 is good. We will use the following consequence of Skolem’s result.

Corollary 22. For f ∈R(Q) the set {ν ≥ 1:∂ν (f)=0} is good. Indeed, for f ∈R(Q)wehavethatdlog(f) is a rational function to which we apply Proposition 21. As the collection of good subsets is stable under ﬁnite intersection, we also have the following.

Corollary 23. For F ∈R(Q)[T ] the set {ν ≥ 1:∂ν (F )=0in Q[T ]} is good. Corollary 23 follows from Corollary 22 applied to the coeﬃcients of F . By abuse of notation, in Corollary 23, we have written ∂ν : R(Q)[T ] → Q[T ] for the map derived from ∂ν : R(Q) ⊂ Λ(Q) → Q by sending T → T .

Proof of Theorem 20. By assumption we have some 0 = H ∈ Z[T1,...,Tn]such that

(62) H(µk([X1]),...,µk([Xn])) = 0 in R(Q)[T ]. N ∈ Z Decompose H = i=1 Hi into a product of irreducible Hi [T1,...,Tn]and consider

(63) Xi := {ν ≥ 1|∂ν (Hi(µk([X1]),...,µk([Xn]))) = 0}, 1 ≤ i ≤ N.

By Corollary 23 the Xi ⊂ N are good and for ﬁxed ν ≥ 1wehave N (62) 0 = ∂ν (H(µk([X1]),...,µk([Xn]))) = ∂ν (Hi(µk([X1]),...,µk([Xn]))) in Q[T ]. i=1 Q N N As [T ]isanintegraldomainthisimpliesthat i=1 Xi = .

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We claim that this implies the existence of 1 ≤ i0 ≤ N and M ≥ 1 such that MN ⊂ Xi . Indeed, assume that this is not the case and write 0 Xi =Σi (j + MiN) ,i=1,...,N,

j∈Ii

for suitable ﬁnite sets Σi, integers Mi ≥ 1 and subsets Ii ⊂{0,...,Mi − 1}.Then ∈ 0 Ii for i =1,...,N . Consider a positive multiple x of the product M1 ...MN ∈ N ∈ ∈ N so large that x Σi.As = Xi we ﬁnd i0 and j Ii0 such that x j + Mi0 ≡ ≡ ∈ hence x j (mod Mi0 )andx 0(modMi0 ), a contradiction because j Ii0 , hence ≡ ≥ N ⊂ j 0(modMi0 ), so there do indeed exist M 1andi0 such that M Xi0 .

We will see that G := Hi0 and M satisfy the requirements of Theorem 20. By construction, G is irreducible, and it remains to be checked that

G(µk ([X1 ×k kM ]),...,µk ([Xn ×k kM ])) = 0. M M R Q → Q Equivalently, as ν≥1 ker(∂ν : ( )[T ] [T ]) = 0 we need to show that for all ν ≥ 1 × × (64) ∂ν (G(µkM ([X1 k kM ]),...,µkM ([Xn k kM ]))) = 0.

Using (19) and ∂ν FM = ∂νM [DG], V, 5, 2.6, we obtain × × = G(∂ν(µkM ([X1 k kM ])),...,∂ν (µkM ([Xn k kM ])))

= G(∂ν(FM (µk([X1]))),...,∂ν (FM (µk([Xn]))))

= ∂νM(G(µk([X1]),...,µk([Xn]))). ∈ N ⊂ However, νM M Xi0 , so this vanishes by construction as G = Hi0 ;see (63).

3.4. Zero-dimensional varieties. Let k be a finite field. Before applying the results of Section 3.3 we include in this subsection a digression on the subring of K0(Vark) generated by zero-dimensional varieties. Generally speaking, the difficulty of the problem of deciding whether the classes of given varieties [X1],...,[Xn] are algebraically independent in K0(Vark)grows quite rapidly with n.Forn = 1 a complete answer can be given. Theorem 24. Let k be a finite field and let X/k be separated and of finite type. Then the following are equivalent:

1) [X] ∈ K0(Vark) is algebraically dependent. 2) [X] is integral over Z. 3) dim (X)=0. Condition 1) means that [X] satisfies some polynomial with coefficients in Z but this polynomial does not have to be monic. So, a priori, condition 2) is stronger than condition 1). Let Z˜ ⊂ K0(Vark) denote the integral closure of Z inside K0(Vark) and let S ⊂ K0(Vark) be the subring generated by the classes of zero-dimensional varieties. By Theorem 24 we have S ⊂ Z˜ and even more, for all X/k,if[X] ∈ Z˜, then [X] ∈ S. Note that this does not imply Z˜ ⊂ S because an element of K0(Vark) is in general only a formal difference of varieties. In fact, I do not know whether S ⊂ Z˜ is an equality. One can give a simple presentation of the ring S; see Theorem 25 below.

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Proof of Theorem 24. We start by showing that 3) implies 2). Let X/k be zero red dimensional, i.e. X = Spec(A) for some Artinian k-algebra A.As[X]=[X ]we | can assume that A is reduced, i.e. A = Li for Li k finite field extensions. Then [X]= [Spec(Li)], and we are reduced to showing that for a finite extension field k ⊂ L we have [Spec(L)] ∈ Z˜.But,ifd =[L : k], then 2 [Spec(L)] − d [Spec(L)] = [Spec(L ⊗k L)] − d [Spec(L)] = [Spec(L⊕d)] − d [Spec(L)] = 0.

⊕d We used that L|k is Galois to have L ⊗k L L as k-algebras. That 2) implies 1) is trivial, and we now show that 1) implies 3). Let X/k be separated and of ﬁnite type and such that [X] is algebraically dependent. We need to show that dim(X)=0. As [X]=[Xred] and dim(X)=dim(Xred) we assume that X is reduced. As k is perfect this implies that the smooth locus of X/k is (open and) dense. This will be used later, but ﬁrst we exploit the assumption that [X]is algebraically dependent. There is 0 = F ∈ Z[T1]withF ([X]) = 0. Let us consider f := µk([X]) ∈R(Q)[T ]. For any ν ≥ 1wehave

F (∂ν (f)) = F (∂ν (µk([X]))) = (∂ν ◦ µk)(F ([X])) = 0. Q d As ∂ν (f) lies in [T ]andF =0,thisimpliesthat∂ν (f) is a constant, i.e. dT (∂ν (f)) =0.Thenalso df d ∂ν ( )= (∂ν (f)) = 0 for all ν ≥ 1 dT dT R Q → Q df and using ν≥1(ker(∂ν : ( )[T ] [T ])) = 0 we get dT = 0, i.e. f itself is a constant, f ∈R(Q). Now assume by contradiction that d := dim (X) ≥ 1. The coefficient of T 2d in f is, by definition of µ , the image under χ : K (Rep Q ) → k q 0 Gk l Q − j W j Λ( l) of the virtual Galois representation j( 1) [Gr2d Hc (X)], and this virtual representation is zero because f is a constant and 2d>0, i.e. − j W j (65) ( 1) [Gr2d Hc (X)] = 0. j From [D3], 3.3.4, we have W j (66) Gr2d Hc (X)=0forj =2d and W 2d (67) Grν Hc (X)=0forν =2d. W 2d From (65) and (66) we get Gr2d Hc (X) = 0 which by (67) implies 2d (68) Hc (X)=0. By the introductory remarks concerning the smooth locus of X/k we can choose U ⊂ X open, smooth and purely d-dimensional such that the dimension of X − U is strictly less than d (U need not be dense in X; it may be chosen as the union of the smooth loci of the irreducible components of X of dimension d). The last con- 2d Q → 2d Q dition assures that the natural map Hc (U, l) Hc (X, l) is an isomorphism, 2d Q 0 Q hence (68) gives Hc (U, l)=0.Poincaré duality gives H (U, l)=0whichisa 0 Q contradiction because dimQl H (U, l) is the number of connected components of U and U = ∅.

We now give a presentation of the subring S ⊂ K0(Vark) generated by zero- dimensional varieties. From the proof of Theorem 24, 3)⇒2), we know that S is

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generated by the classes of ﬁnite ﬁeld extensions of k. Putting xν := [Spec(kν )] ∈ K0(Vark)(ν ≥ 1) we have

S = Z[xν : ν ≥ 1] ⊂ K0(Vark). For µ, ν ≥ 1wewrited(µ, ν)(resp. m(µ, ν)) for the greatest common divisor (resp. the least common multiple) of µ and ν. Computing tensor products of finite extensions of finite fields, one checks that xµxν = d(µ, ν)xm(µ,ν),andwegeta surjective ring homomorphism

(69) φ : Z[Tν : ν ≥ 1]/(Tν Tµ − d(µ, ν)Tm(µ,ν) : µ, ν ≥ 1) −→ S, Tν → xν .

Theorem 25. The above φ is an isomorphism. In particular, the xν (ν ≥ 2) are pairwise distinct zero divisors in K0(Vark).

Remark. The existence of zero divisors in K0(Vark) was first addressed by B. Poo- nen in [P] where he showed their existence for k of characteristic zero. One can show that for any finitely generated field k the images of the classes of the non- A1 −1 trivial finite Galois extensions of k in K0(Vark) k] ] are an infinite set of zero divisors.

Proof of Theorem 25. The left-hand side of (69) is a free Z-module on the set {Tν : ν ≥ 1}. To show that φ is injective we use that, for all n ≥ 1, sending [X] →|X(kn)| deﬁnes a motivic measure ψn : K0(Vark) → Z and that, writing Xν := Spec(kν ), we have

|Xν (kn)| = ν,ifν divides n,

|Xν (kn)| =0,otherwise, ≥ for all ν, n 1. Now assume that φ( ν≥1 aν Tν )= ν≥1 aν [Xν ]=0forsome aν ∈ Z, almost all zero. We need to show that aν =0forallν. For any n ≥ 1we have (∗n)0=ψn( aν [Xν ]) = aν |Xν (kn)| = aν ν. ν≥1 ν≥1 ν|n

To show that all aν are zero we proceed by induction on the number of prime divisors, counting multiplicity, of ν. ∗ For ν =1wehave( )1 : a1 =0. ∗ For ν =1wehavefrom()ν that d|ν add = 0. By induction hypothesis we have ad = 0 for all proper divisors d of ν, hence aν ν =0andaν =0. This shows that φ is injective, hence an isomorphism. To prove the last claim of Theorem 25 we use that, since φ is an isomorphism, S is Z-free on the set {xν : ν ≥ 1} andinparticularthatthexν are all distinct. Finally, for ν ≥ 2, 2 − − − − 0=xν νxν = xν (xν ν). As xν ν = xν νx1 =0(notethatx1 =1)wesee that xν is a zero divisor. Remark. The above considerations suggest a simple example illustrating Theorem 20. We take n := 1, X1 := Spec(k2) in this result and compute µk([X1]) ∈R(Q)[T ]. i 0 We have Hc(X1)=0fori =0andHc (X1) is two dimensional with geometric Frobenius Fk acting by exchanging the vectors of a (ﬁxed) basis. This gives for ν ≥ 1 ν | 0 (70) cν := tr(Fk Hc (X1)) = 2 for ν even

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and cν =0forν odd. For arbitrary G ∈ Z[T1]wehaveG(µk([X1])) = 0 if and only if G(∂ν(µk([X1]))) = 0 for all ν ≥ 1, and from the above computation of the cohomology of X1 and Proposition 8i) we get that ∂ν (µk([X1])) ∈ Q[T ]isinfact a constant equal to cν .SoG(µk([X1])) = 0 is equivalent to G(cν ) = 0 for all ν ≥ 1. From (70) we see that the assertion of Theorem 20 holds with M =2and G = T1 − 2. However, there is no irreducible polynomial G ∈ Z[T1] satisfying G(µk([X1])) = 0 because such a G must, by (70), satisfy G(0) = G(2) = 0 and hence must be divisible by T (T − 2). This shows that in Theorem 20 one has to allow for a finite extension of the base field in order to find an irreducible relation. 3.5. Two curves. Let k be a finite field. In this subsection by a curve over k we shall always mean a proper, smooth and geometrically connected curve. We are going to treat in detail the following special case of the problem of algebraic independence in K0(Vark). Let X and Y be curves over k.Whenare[X]and [Y ] algebraically independent in K0(Vark), i.e. when is the subring Z [[X], [Y ]] ⊂ K0(Vark) a polynomial ring in the variables [X]and[Y ]? We will see that this is the case for a generic pair (X,Y ). First we define special curves which will turn out to be those for which our methods can shed no light on the above question. A curve X/k is special if i) 2n the degree of k over its prime field is even: k = Fq, q = p (n ≥ 1) and ii) all 1 eigenvalues of the geometric Frobenius attached to k acting on Hc (X)areequal to +q1/2.Notethatbyi)+q1/2 is a rational integer so ii) is meaningful. Being special is not a property of the curve alone but depends on the base field, too. The projective line is special over any k satisfying condition i). An elliptic curve X/k becomes special after a finite extension of the base field if and only if it is super- singular. If X/k is special, then the Jacobian of X is k-isogeneous to the power of a super-singular elliptic curve, and we see that most curves of genus at least one are non-special. The main result of this subsection is the following.

Theorem 26. Let X1 and X2 be proper, smooth and geometrically connected curves over the finite field k such that [X1] and [X2] are algebraically dependent in K0(Vark). Then there is a finite extension k ⊂ K such that at least one of the following holds: 1) Both X1 ×k K/K and X2 ×k K/K are special. 2) The Jacobians of X1 and X2 are K-isogeneous. Corollary 27. If E/k is an ordinary elliptic curve, then [P1] and [E] are algebraically independent in K0(Vark). This is clear using Theorem 26 and the discussion preceding it. Theorem 26 can be applied to the following problem: It is difficult to exhibit relations in K0(Vark). One such relation, which follows directly from the definition, is that if X −→ Y is a Zariski-locally trivial fibre bundle with fibre G,then[X]=[G][Y ]inK0(Vark). One might hope for a similar relation for more general X −→ Y ,e.g.étale covers. However: Corollary 28. Let X/k be a curve of genus at least two which does not become special after any finite extension of k.IfY −→ X is a non-trivial finite étale cover, then [X] and [Y ] are algebraically independent in K0(Vark).

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Proof. By the assumption made on X, condition 1) of Theorem 26 is excluded. The Hurwitz formula implies that the genus of Y is strictly larger than the genus of X, excluding condition 2) of Theorem 26. Hence Theorem 26 implies the algebraic independence of [X]and[Y ].

Remark. Rephrasing Theorem 26 we have: If for two given curves X1,X2/k none of 1) or 2) from Theorem 26 holds after any finite extension of k,then[X1]and[X2] are algebraically independent. We will in fact show that in this situation µk([X1]) and µk([X2]) are algebraically independent in R(Q)[T ]. We have seen in Corollary 3 1 that for the algebraic independence of the µk([Xi]) it is sufficient that [Hc (X1)] and 1 Q [Hc (X2)] are algebraically independent in K0(Rep Gk l). The following example shows that Theorem 26 is sharper than just the algebraic independence of H1’s. Let X2/k be a curve such that its Jacobian is k-isogeneous to the square of an ordinary elliptic curve X1/k (such a curve exists for suitable k by [BDS], Theorem 1 1 ⊕2 1 1 2). Then Hc (X2) Hc (X1) as Gk-modules and so [Hc (X1)] and [Hc (X2)] are Q algebraically dependent in K0(Rep Gk l). But Theorem 26 still shows that the µk([Xi]) (i =1, 2) (and hence the [Xi] themselves) are algebraically independent. The curve X1 ×k K/K is not special for any finite extension k ⊂ K as explained before Theorem 26 and the Jacobians of X1 and X2 are not isogeneous because they are of different dimensions. Another consequence of Theorem 26 is that two curves of different genera have algebraically independent classes unless they both become special after some finite extension of the base field. As the referee points out, over k = C one can easily show that two curves of different genera have algebraically independent classes using a motivic measure based on the Hodge polynomial; see [LL2], 3.4. Proof of Theorem 26. Since we are allowed to make a finite extension of k we can assume by Theorem 20 that there is an irreducible G ∈ Z[T1,T2]with G(µk([X1]),µk([X2])) = 0. Writing

fi,ν := ∂ν (µk([Xi])) ∈ Q[T ],i=1, 2,ν ≥ 1, we have

(71) G(f1,ν ,f2,ν ) = 0 for all ν ≥ 1. Introducing, for ν ≥ 1, − ν | 1 aν := tr(Fk Hc (X1)), − ν | 1 bν := tr(Fk Hc (X2)),

where Fk is the geometric Frobenius of k = Fq, say, we have ν 2 f1,ν =1+aν T + q T , ν 2 f2,ν =1+bν T + q T 0 Q 2 Q − because Hc (Xi)= l and Hc (Xi)= l( 1) and using Proposition 8i). We consider 2 2 (72) F (T1,T2):=q(T2 − T1) − (a1 − b1)(a1T2 − b1T1)+(a1 − b1) .

The discriminant of F , considered as a polynomial in T2,equals − 2 − 2 2 − (73) 4q(a1 b1) T1 +(a1 b1) (a1 4q).

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The choice of F is made such that we have

2 2 (74) F (f1,1,f2,1)=F (1 + a1T + qT , 1+b1T + qT ) 2 2 2 = q(b1 − a1) T − (a1 − b1)(a1 + a1b1T + a1qT 2 2 −b1 − b1a1T − b1qT )+(a1 − b1) 2 2 2 2 2 = q(b1 − a1) T − (a1 − b1) (1 + qT )+(a1 − b1) 2 2 2 2 =(a1 − b1) (qT − 1 − qT )+(a1 − b1) =0.

To continue, we make the following additional assumption: 1) a1 = b1: The prime ideal

P1 := ker (Q[T1,T2] −→ Q[T ],Ti → fi,1)

is proper and non-zero, hence of height 1 in Q[T1,T2]. By (71) 0 = G ∈ P1 and G is irreducible. Since Q[T1,T2] is factorial we have P1 =(G), the principal ideal generated by G.Asa1 = b1 the discriminant (73) of F is not a square in Q[T1], so F is irreducible in Q[T1,T2]. By (74) we have F ∈ P1, and the same reasoning as ∗ for G gives P1 =(F )=(G). So there is some α ∈ Q with F = αG from which we (71) get F (f1,ν ,f2,ν )=αG(f1,ν ,f2,ν ) =0forallν ≥ 1 which we make more explicit:

ν 2 ν 2 0=F (f1,ν ,f2,ν )=F (1 + aν T + q T , 1+bν T + q T ) (72) 2 = q((bν − aν )T ) − (a1 − b1) ν 2 ν 2 2 (a1 + a1bν T + a1q T − b1 − b1aν T − b1q T )+(a1 − b1) 2 2 ν ν = T (q(bν − aν ) − (a1 − b1)(a1q − b1q )) + T (−(a1 − b1) 2 2 (a1bν − b1aν )) − (a1 − b1) +(a1 − b1) 2 2 ν 2 = T (q(bν − aν ) − q (a1 − b1) )+T ((a1 − b1)(b1aν − a1bν )).

So for any ν ≥ 1(usinga1 − b1 =0):

2 ν 2 (75) q(bν − aν ) = q (a1 − b1) ,

(76) b1aν = a1bν .

Now we assume in addition to a1 = b1 that we also have 1.1) a1 =0and b1 =0:From(75)wehave 2 − 2 ν−1 − 2 bν 2aν bν + aν = q (a1 b1) into which we substitute (76) to get

2 2 b1 − 2 b1 2 ν−1 − 2 aν 2 2aν + aν = q (a1 b1) . a1 a1

2 Multiplying this by a1 gives 2 2 − 2 ν−1 − 2 2 aν (b1 2a1b1 + a1)=q (a1 b1) a1,

2 and cancelling (a1 − b1) =0weobtain 2 ν−1 2 ≥ (77) aν = q a1 for ν 1.

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We have a2 =tr(F ν |H1(X )⊗2), hence ν k c 1 ⎛ ⎞ ν − | 1 ⊗2 ⎝− 2 t ⎠ (78) det(1 Fkt Hc (X1) )=exp aν ≥ ν ⎛ ⎞ ⎛ ν 1 ⎛ ⎞⎞ ν ν (77) t (qt) =exp⎝− qν−1a2 ⎠ =exp⎝q−1a2 ⎝ − ⎠⎠ 1 ν 1 ν ν≥1 ν≥1 a2/q =(1− qt) 1 .

The ﬁrst equality in (78) is [H], Appendix C, Lemma 4.1. We denote by g(Xi)the 1 genus of Xi.AsdimQl Hc (Xi)=2g(Xi) we obtain from (78) by comparing degrees: 2 2 (79) a1 =4g(X1) q. 1 Denoting by α1,...,α2g(X1) the eigenvalues of Fk acting on Hc (X1)consideredas ∼ 1/2 complex numbers via some isomorphism Ql → C,wehave|αi| = q and thus

2g(X1) | 2| |− |2 ≤ 1/2 2 2 a1 = αi (2g(X1)q ) =4g(X1) q. i=1

By (79) this upper bound is in fact an equality which forces all αi to be equal to either +q1/2 or −q1/2. After possibly a quadratic extension of k we will have all 1/2 αi equal to +q and then X1/k will be special by definition. As the additional assumptions we have made so far, namely that a1 = b1 and a1,b1 = 0, are symmetric in X1 and X2 the same conclusion holds for X2/k, and we are in case 1) of Theorem 26. Now we assume: 1.2) a1 = 0: We then have, as we still assume that a1 = b1,thatb1 =0and ≥ 1 P1 (76) implies that aν =0forallν 1, hence Hc (X1)=0andX1 .Notethat there are no non-trivial k/k-forms of P1 because the Brauer group of a finite field 2 ν−1 2 is trivial. From (75) we then get bν = q b1,andthesameargumentasincase 1.1) (cf. (77)) shows that X2 becomes special after at most a quadratic extension. So we are again in case 1) of Theorem 26. 1.3) b1 = 0: this case is symmetric to the case 1.2) above. There is one case left to be considered. 2) a1 = b1:Wenowhavef1,1 = f2,1 and F := T1 − T2 satisfies F (f1,1,f2,1)=0. By the same arguments as in case 1) we get F (f1,ν ,f2,ν ) = 0 for all ν ≥ 1, i.e. ν 2 ν 2 0=F (1 + aν T + q T , 1+bν T + q T )=(aν − bν )T,

so aν = bν for all ν ≥ 1fromwhichweget − | 1 − | 1 det(1 Fkt Hc (X1)) = det(1 Fkt Hc (X2))

which is equivalent to the Jacobians of X1 and X2 being k-isogeneous [T], Theorem 1, c1)⇔c2), and hence we are in case 2) of Theorem 26. Remark. We conserve the notations of the above proof. The simplest case in which Theorem 26 does not guarantee the algebraic independence of the classes of two 1 curves in K0(Vark)isifX1 = P and X2 = E is a super-singular elliptic curve. Let 2 k be so large that X2/k is special and put q := |k|.ThenF := q(T2−T1) −4qT1+4q satisﬁes

(80) F (µk([X1]),µk([X2])) = 0.

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In fact, we have ν 2 f1,ν =1+q T = ∂ν (µk([X1])), ν/2 ν 2 f2,ν =1− 2q T + q T = ∂ν (µk([X2])) and compute ν/2 2 ν 2 F (∂ν (µk([X1])),∂ν(µk([X2]))) = q(−2q T ) − 4q(1 + q T )+4q =4qν+1T 2 − 4qν+1T 2 − 4q +4q =0, for all ν ≥ 1, hence (80) follows. On the other hand + − F ([X1], [X2]) = [Z ] − [Z ]inK0(Vark), where Z+ and Z− are the (non-connected) proper, smooth varieties Z+ =(E2) q ((P1)2) q Spec(k) 4q, Z− =(E × P1) 2q (P1) 4q. The sign means disjoint union of schemes. I cannot decide whether [Z+]and − [Z ]areequal(i.e.whetherF ([X1], [X2]) = 0), but I suspect not and even more, that [P1]and[E] should be algebraically independent. Note at least that Z+ Z−, e.g. Z+ has isolated points but Z− does not. This illustrates that our methods can only give sufficient conditions for varieties to be algebraically independent in the Grothendieck ring of varieties, but we have no idea how to produce interesting relations in K0(Vark). Acknowledgements I would like to thank my advisor C. Deninger for constant support and encour- agement during the preparation of my thesis, the main results of which are pre- sented here. Furthermore, I would like to thank H. Frommer, U. Jannsen, M. Kisin, B. Moonen, D. Roessler, C. Serpé and C. Soulé for useful discussions and the referee for a very thorough report. References

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NWF I- Mathematik, Universitat¨ Regensburg, Universitatsstrasse¨ 31, 93053 Regens- burg, Germany

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